Question

Find the equation of the circle when the end points of a diameter are $A(3,2)$ and $B(3,5)$

Answer 1

Using diameter form,the equation of the circle having $A(3,2)$ and $B(3,5)$ as the end points of a diameter is

$(x-3)(x-3)+(y-3)(y-2)=0$

$\Rightarrow x^2-3x-3x+9+y^2-3y-2y+6=0$

$\Rightarrow x^2+y^2-6x-5y+15=0$ is the required equation of the circle.